
%!TEX program = xelatex
%!TEX TS-program = xelatex
%!TEX encoding = UTF-8 Unicode

\documentclass[10pt]{article} 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 一些常用的包总结在另一个文件里
\input{wang_preamble.tex}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 选择Windows操作系统写中文文档，使用 xelatex 或 lualatex 编译器
%\usepackage{xeCJK} % 处理中文、日文和韩文（统称为 CJK 文字）的排版
%\setCJKmainfont{SimSun} % 设置正文字体为宋体
%\setCJKmonofont{SimHei} % 设置加粗字体为黑体
%\setCJKsansfont{SimHei} % 黑体

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%选择Mac操作系统写中文文档，使用 xelatex 或 lualatex 编译器
\usepackage{xeCJK} % 支持中文字体
\setCJKmainfont{Songti SC} % 设置主要中文字体，用于正文中的中文文本。设置主要中文字体为宋体
%\setCJKmonofont{Menlo} % 设置等宽中文字体，用于代码块、等宽文本等。设置等宽中文字体为 Menlo
%\setCJKsansfont{PingFang SC} % 设置无衬线中文字体，用于标题、图表标签等。设置无衬线中文字体为 PingFang SC
%\setCJKromanfont{Songti SC} % 设置罗马中文字体，用于罗马字体中的中文文本。设置罗马中文字体为宋体

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%选择输出文档的三种类型：
%\newcommand{\showsolution}{0} %%设置showsolution=0, 编译生成试卷
%\newcommand{\showsolution}{1} %%设置showsolution=1, 编译生成试卷解答、考完发给学生
\newcommand{\showsolution}{2} %%设置showsolution=2, 编译生成试卷解答与评分标准、阅卷与试卷袋归档

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 填写课程信息：
\newcommand{\CourseName}{复变函数}
\newcommand{\CourseNumber}{162250220}
\newcommand{\CourseStudents}{2022级数学与应用数学}
\newcommand{\CourseTerm}{2024 $\sim$ 2025 学年 第 一 学期}
\newcommand{\ExamAB}{B}
\newcommand{\ExamContents}{本次考试的主要内容是：
复数的代数运算、
用复数方程表示平面几何图形、
计算有理函数的部分分式、
计算指数函数、
计算对数函数、
使用交比找出符合条件的线性变换、
计算复积分、
使用幅角原理计算解析函数的零点、
使用留数定理计算实积分、
将亚纯函数展开成幂级数。
%考试内容覆盖教学大纲的大部分重要内容。
}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%考完发给学生：
\ifnum\showsolution=1

\usepackage{titling}
\setlength{\droptitle}{-2cm}   % 标题上移2cm

\author{王立庆（\CourseStudents）}
\title{\CourseName 考试 \ExamAB 解答 }
\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
\date{2024年12月19日}

\fi

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 试卷：校训、课程信息、姓名学号
\ifnum\showsolution=0

\begin{center}
\includegraphics [width=0.85\textwidth, height=4.1cm]{lixin-pan-new.eps}
\end{center}

\vspace{-0.5cm}

\begin{center}
{\Large \bf 上海立信会计金融学院期终考试卷  \hspace{0.3cm} \underline{\,\ExamAB\, } 卷 }

\vspace{0.3cm}

{\large \CourseTerm }

\vspace{0.3cm}

{\large \bf \underline{\,\CourseStudents\,} 《\underline{\,\CourseName\,}》 课程代码：\underline{\,\CourseNumber\,} }

\vspace{0.3cm}

（本场考试属\underline{ \, 闭 \, }卷考试，考试时间共\underline{ \, 90 \,  }分钟，不准使用计算器）共\underline{ \, \pageref{LastPage} \, }页 

\vspace{0.7cm}

班级 \underline{\hspace{3.5cm}} 学号 \underline{\hspace{3.5cm}} 姓名 \underline{\hspace{3.5cm}} 

\end{center}

\vspace{-0.2cm}

\begin{table}[h]
\centering
\renewcommand{\arraystretch}{1.2}
\begin{tabular}{
|>{\centering\arraybackslash}p{0.8cm}|>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}|
>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}|
>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}|
>{\centering\arraybackslash}p{0.7cm}|>{\centering\arraybackslash}p{0.7cm}
|>{\centering\arraybackslash}p{1.0cm}|>{\centering\arraybackslash}p{1.2cm}|>{\centering\arraybackslash}p{1.2cm}|}
\hline
题号 &一&二&三&四&五&六&七&八&九&十&总分&合成人签名&审核人签名 \\
\hline
得分 $\,\,\,\,\,\,\,\,$ &&&&&&&&&&&&& \\
\hline
\end{tabular}
\end{table}

\vspace{0.2cm}

本次考试共10题，每题10分。

\fi

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 考完发给学生：给文档加标题、作者、日期、摘要
\ifnum\showsolution=1

\maketitle

\abstract{\ExamContents }

\vspace{1cm}

\thispagestyle{fancy} % 第一页也显示“第几页，共几页”的信息。

\fi

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 解答与评分标准：校名、课程信息
\ifnum\showsolution=2

\begin{center}

{\Large \bf 上海立信会计金融学院期终考试卷 \,\, \underline{\, \ExamAB \, } 卷\,\, 解答与评分标准}

\vspace{0.3cm}

{\large \CourseTerm }

\vspace{0.3cm}

{\large \bf \underline{ \, \CourseStudents \, } 《\underline{ \, \CourseName \, }》 课程代码：\underline{ \, \CourseNumber \, }  }

\end{center}

\vspace{0.2cm}

本次考试共10题，每题10分。

\vspace{0.2cm}

\fi

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 题目和解答从这里开始

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item %1
Find the value of
$$(1+2i)^2 + \frac{25}{3+4i} + \sin (6i).$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 试卷留空位
\ifnum\showsolution=0
\vspace{3cm}
\newpage
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第1题：解答与评分标准
\ifnum\showsolution>0

{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  By arithmetic rules of complex numbers, and the definition of the sine function, we obtain 
\begin{equation*}
\begin{aligned}
(1+2i)^2 + \frac{25}{3+4i} + \sin (6i) 
&= 1+2(2i)+(2i)^2 + \frac{25(3-4i)}{(3+4i)(3-4i)} + \frac{e^{i(6i)}-e^{-i(6i)}}{2i} \\ 
& = 1 + 4i -4 + 3 - 4i -\frac{i}{2}(e^{-6}-e^6) \\
& = \frac{e^6-e^{-6}}{2}i. 
\end{aligned}
\end{equation*}

\ifnum\showsolution=2
\dotfill (\underline{\,\,三部分每部分3分，总和1分\,\,})
\fi


\end{enumerate}

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第1题：解答结束
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item %2
When does $az + b\bar{z} + c = 0$ represent a line?

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 试卷留空位
\ifnum\showsolution=0
\vspace{6cm}
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第2题：解答与评分标准
\ifnum\showsolution>0

{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  Let the line be represented by $z=z_0+tz_1$, where $z_0,z_1$ are two fixed complex numbers and $z_1\neq 0$. 
The parameter $t$ can be take any real number. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  Replace $z$ in $az + b\bar{z} + c = 0$ by $z=z_0+tz_1$, we get 
$ a(z_0+tz_1) + b(\bar{z}_0+t\bar{z}_1) + c =0. $

\item  Since $t$ can take any real values, we must have 
\begin{equation*}
\begin{aligned}
az_0 + b\bar{z}_0+c &=0, \\
az_1+b\bar{z}_1 &=0. 
\end{aligned}
\end{equation*}
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  From $az_1+b\bar{z}_1=0$ we get $(a\bar{a} - b\bar{b})z_1=0$; Since $z_1\neq 0$, we get $a\bar{a}=b\bar{b}. $
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  By the existence of $z_0$ in $az_0 + b\bar{z}_0+c =0$, we consider its conjugate $\bar{a}\bar{z}_0 + \bar{b}z_0+\bar{c} =0$ and then eliminate $\bar{z}_0$,  hence we get $z_0 = \frac{b\bar{c}-c\bar{a}}{a\bar{a}-b\bar{b}}. $ 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  Hence we arrive at the conditions $a\bar{a}=b\bar{b}$ and $b\bar{c}=c\bar{a}$. 

\item  When $a=1$, $b=e^{2i\theta}$, $c=re^{i\theta}$, we see that the equation $z+e^{2i\theta}\bar{z}+re^{i\theta} =0$ can be written as $e^{-i\theta}z + e^{i\theta}\bar{z} = -r$. Denote $w=e^{-i\theta}z$, and we get $w+\bar{w}=-r$. This is the line $\mathrm{Re}(w)=-\frac{r}{2}$. 

\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi


\end{enumerate}

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第2题：解答结束
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item %3
Suppose the derivative of a real function of a complex variable exists. Prove that the derivative is zero. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 试卷留空位
\ifnum\showsolution=0
\vspace{6cm}

\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第3题：解答与评分标准
\ifnum\showsolution>0

{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  Let $f(z)$ be a complex variable function, and it takes only real values. 

\item  Suppose $f'(z)$ exists. That is, the limit $\lim\limits_{h\to 0} \frac{f(z+h)-f(z)}{h}$ exists. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  Let $h=\Delta x$ tends to zero through real values, we see that the limit $f'(z)$ is real. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,3分\,\,})
\fi

\item  Let $h=i\Delta y$ tends to zero through purely imaginary values, we see that limit $f'(z)$ is purely imaginary. 

\ifnum\showsolution=2
\dotfill (\underline{\,\,3分\,\,})
\fi

\item  Thus the derivative can only be zero. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\end{enumerate}

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第3题：解答结束
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item %4
Define the exponential function $e^z$ as the solution of the differential equation
$f'(z) = f(z)$ with the initial value $f(0)=1$. Show that $e^z$ satisfies the addition theorem
$e^{a+b} = e^a\cdot e^b$. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 试卷留空位
\ifnum\showsolution=0
\vspace{6cm}
\newpage 
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第4题：解答与评分标准
\ifnum\showsolution>0

{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  Let $c$ be a constant complex number. Let $g(z)=f(z)f(c-z)$. 

\item  By the product rule of derivative, and by the condition $f'(z) = f(z)$, we have 
\begin{equation*}
\begin{aligned}
g'(z) &= f'(z)f(c-z)+f(z)f'(c-z)(-1) \\ 
& = f(z)f(c-z)-f(z)f(c-z) \\ 
& =0.
\end{aligned}
\end{equation*}
\ifnum\showsolution=2
\dotfill (\underline{\,\,3分\,\,})
\fi

\item  Since the derivative of $g(z)$ is zero, we see that $g(z)=g(0)$, it is a constant function.  
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  By the other condition $f(0)=1$, we obtain $f(z)f(c-z)=f(0)f(c)=f(c).$
\ifnum\showsolution=2
\dotfill (\underline{\,\,3分\,\,})
\fi

\item  Replace $z=a$ and $c-a=b$, we obtain $f(a)f(b)=f(a+b).$
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi



\end{enumerate}

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第4题：解答结束
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item %5
Let $w=f(z)$ be an analytic function in the region $\Omega$. Let $z_0\in \Omega$. 
Prove that the linear change of scale at $z_0$, effected by the transformation $w = f(z)$, is independent of the direction. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 试卷留空位
\ifnum\showsolution=0
\vspace{6cm}
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第5题：解答与评分标准
\ifnum\showsolution>0

{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  Since $f(z)$ is analytic in the region $\Omega$ and $z_0\in \Omega$, the derivative $f'(z_0)$ exist. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  The derivative is defined by the existence of the limit $$f'(z_0) = \lim\limits_{z\to z_0} \frac{f(z)-f(z_0)}{z-z_0}. $$
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  Taking modulus, we have $$|f'(z_0)| = \lim\limits_{z\to z_0} \frac{ |f(z)-f(z_0)| }{ |z-z_0| }. $$
\ifnum\showsolution=2
\dotfill (\underline{\,\,3分\,\,})
\fi
\item  This means that $|f(z)-f(z_0)| = |f'(z_0)|\cdot |z-z_0| + o(|z-z_0|)$.  

\item  Consider any small line segment $z-z_0$ with one end point at $z_0$, and it image by the mapping $w=f(z)$. 

\item  The scale of these two segments is, in the limit, contracted or expanded in the ratio $|f'(z_0)|$, and it is independent of the direction of $z-z_0$. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,3分\,\,})
\fi


\end{enumerate}

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第5题：解答结束
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item %6
Explain that the linear transformation 
\begin{equation*}
w=\frac{az+b}{cz+d}
\end{equation*}
is composed by a translation, an inversion, a rotation, and a homothetic transformation followed by another translation. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 试卷留空位
\ifnum\showsolution=0
\vspace{6cm}
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第6题：解答与评分标准
\ifnum\showsolution>0

{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  When $c=0$, the linear transformation becomes $$w=\frac{a}{d}z + \frac{b}{d} = re^{i\theta}z + \frac{b}{d}, $$
where $\frac{a}{d} = re^{i\theta}$. It is the composition of a rotation $z\mapsto e^{i\theta}z$, a homothetic transformation $u\mapsto ru$ and a translation $v\mapsto v+\frac{b}{d}$. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  When $c\neq 0$, the linear transformation becomes 
$$w = \frac{az+b}{cz+d} = \frac{a(z+\frac{d}{c})}{c(z+\frac{d}{c})} + \frac{b-\frac{ad}{c}}{cz+d}
= \frac{a}{c} + \frac{bc-ad}{c^2(z+\frac{d}{c})}. $$
\ifnum\showsolution=2
\dotfill (\underline{\,\,3分\,\,})
\fi

\item  Let $\frac{bc-ad}{c^2}=re^{i\theta}$.
The linear transformation is composed by a translation $z\mapsto z+\frac{d}{c}$, an inversion $u\mapsto \frac{1}{u}$, a rotation $v\mapsto e^{i\theta}v$, a homothety $\xi\mapsto r\xi$, and another translation $\eta\mapsto \eta+\frac{a}{c}$. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,5分\,\,})
\fi


\end{enumerate}

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第6题：解答结束
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item %7
Compute $$\int_{|z|=1} \frac{e^zdz}{z}.$$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 试卷留空位
\ifnum\showsolution=0
\vspace{6cm}
\newpage 
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第7题：解答与评分标准
\ifnum\showsolution>0

{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  Consider $f(z)=e^z$ and $a=0$. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  By Cauchy's integral formula, we have 
$$
\int_{|z|=1} \frac{e^zdz}{z} 
= \int_{|z|=1} \frac{f(z)dz}{z-0}
= f(0)\cdot 2\pi i \cdot n(\gamma, 0)
=e^0\cdot 2\pi i\cdot 1
= 2\pi i. 
$$

\ifnum\showsolution=2
\dotfill (\underline{\,\,8分\,\,})
\fi


\end{enumerate}

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第7题：解答结束
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item %8
Suppose that $\varphi(\zeta)$ is continuous on the arc $\gamma$. 
Define a sequence of functions 
$$
F_n(z) = \int_\gamma \frac{\varphi(\zeta)d\zeta}{(\zeta-z)^n}. 
$$
Prove that $F_1(z)$ is analytic in each of the regions determined by $\gamma$, and its derivative is 
$F'_1(z) = F_2(z)$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 试卷留空位
\ifnum\showsolution=0
\vspace{6cm}
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第8题：解答与评分标准
\ifnum\showsolution>0

{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  We first prove that $F_1(z)$ is continuous. 

\item  Let $z_0$ be  a point in one of the regions determined by $\gamma$. 

\item  Since $\gamma$ is a compact set, we choose the neighbourhood $|z-z_0|<\delta$ so that it does not meet $\gamma$. 

\item  By restricting $z$ to the smaller neighbourhood $|z-z_0|<\frac{\delta}{2}$ we have $|\zeta-z|>\frac{\delta}{2}$ for all $\zeta\in\gamma$. 

\item  Since $\varphi(\zeta)$ is continuous on the compact set $\gamma$, it has a maximum $M$. We obtain 
\begin{equation*}
\begin{aligned}
|F_1(z)-F_1(z_0)| 
&= \left\vert \int_\gamma \frac{\varphi(\zeta)d\zeta}{\zeta-z} - \int_\gamma \frac{\varphi(\zeta)d\zeta}{\zeta-z_0} \right\vert 
= \left\vert (z-z_0) \int_\gamma \frac{\varphi(\zeta)d\zeta}{(\zeta-z)(\zeta-z_0)} \right\vert  \\ 
& \le |z-z_0| \int_\gamma \frac{2M}{\delta^2}  |d\zeta | . 
\end{aligned}
\end{equation*}
This proves that $F_1(z)$ is continuous at $z_0$. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,4分\,\,})
\fi

\item  We have the quotient of differences
\begin{equation*}
\begin{aligned}
\frac{F_1(z)-F_1(z_0)}{z-z_0} = \int_\gamma \frac{\varphi(\zeta)d\zeta}{(\zeta-z)(\zeta-z_0)}. 
\end{aligned}
\end{equation*}
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  Since $\frac{\varphi(\zeta)}{\zeta-z_0}$ is continuous on $\gamma$, we use the method above again, and see that the integral on the right hand side is continuous for $z$ near $z_0$. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  Taking $z\to z_0$, we obtain 
\begin{equation*}
\begin{aligned}
F_1'(z_0) = \lim\limits_{z\to z_0} \frac{F_1(z)-F_1(z_0)}{z-z_0} 
= \lim\limits_{z\to z_0} \int_\gamma \frac{\varphi(\zeta)d\zeta}{(\zeta-z)(\zeta-z_0)}
= \int_\gamma \frac{\varphi(\zeta)d\zeta}{(\zeta-z_0)^2} 
= F_2(z_0). 
\end{aligned}
\end{equation*}
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\end{enumerate}

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第8题：解答结束
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item %9
Compute the definite integral $$\int_0^\pi \log\sin\theta d\theta. $$ 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 试卷留空位
\ifnum\showsolution=0
\vspace{6cm}
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第9题：解答与评分标准
\ifnum\showsolution>0

{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  The idea is to compute the complex integral for a branch of the logarithm function, 
$$\log\sin z = \log \frac{e^{iz}-e^{-iz}}{2i} = \log \frac{1-e^{2iz}}{-2ie^{iz}} 
=  \log(1-e^{2iz}) - \log(2) -\log(-i) - iz. 
$$
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  The integral of the first term is completed by a contour integral 
$$\int_\gamma \log(1-e^{2iz})dz,$$
where $\gamma$ is the boundary of a rectangle with four cvertices $0,\pi, \pi + iY, iY$, but the vertices $0$ and $\pi$ are avoided by small quadrants since the function has singularities at $z=0$ and $z=\pi$. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  The integrals over the vertical sides cancel each other; 
The integral over the upper horizontal side tends to zero; 
The integrals over the circular quadrants tends to zero as their radius tends to zero; 
The contour integral is zero since the integrand function is analytic inside $\gamma$. 
 \ifnum\showsolution=2
\dotfill (\underline{\,\,4分\,\,})
\fi

\item  The integral of the rest terms is 
$$\int_0^\pi [- \log(2) -\log(-i) - iz] dz = -\pi\log(2)-\pi(-\frac{\pi i}{2}) - i\frac{\pi^2}{2} = - \pi\log(2).$$ 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi


\end{enumerate}

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第9题：解答结束
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\newpage 
\item %10
How many roots does the equation 
$z^5 + z^4 + z^3 + 6z^2 + z + 1 = 0$ 
have in the disk $|z| < 1$? 
Hint: Look for the biggest term when $|z|=1$ and apply Rouche's theorem. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% 试卷留空位
\ifnum\showsolution=0
\vspace{6cm}
\fi

\vspace{0.2cm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第10题：解答与评分标准
\ifnum\showsolution>0

{\color{red}解答：

\begin{enumerate}[label={\arabic*.}]

\item  Consider $f(z)=z^5 + z^4 + z^3 + 6z^2 + z + 1$ and $g(z)=6z^2$. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi

\item  For $|z|=1$, we have 
\begin{equation*}
\begin{aligned}
|f(z)-g(z)| &= |z^5 + z^4 + z^3 + z + 1|\le |z^5| +|z^4|+|z^3|+|z|+1=5, \\  
g(z) &= |6z^2| =6.
\end{aligned}
\end{equation*}
\ifnum\showsolution=2
\dotfill (\underline{\,\,3分\,\,})
\fi

\item  Thus $|f(z)-g(z)| < |g(z)| $ on $|z|=1$. By Rouche's theorem, $f(z)$ and $g(z)$ have the same number of zeros in the unit disk. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,3分\,\,})
\fi

\item  Since $g(z)$ have 2 zeros in the unit disk, we see that $f(z)$ also have 2 zeros. 
\ifnum\showsolution=2
\dotfill (\underline{\,\,2分\,\,})
\fi


\end{enumerate}

}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%第10题：解答结束
\fi

\vspace{0.2cm}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%试卷结束
\end{document}



